Search found 140 matches

Hi, Use isset() at the first line of php code like this and assign the name to submit button : <form name="form1" method="post"> <input type="text" name="name"> <input type="submit" value="login" name="submit"> </form> <?php if(isset($_POST['submit'])) { $name = strtolower($_POST['name']); if ($name...

hi, Declaration of all variables twice is not the correct method, so the easiest way is to declare only once, for example the code should be like this : in html form, include the name of submit button, <input type="submit" name="submit" value="Submit"> <?php if(isset($_POST['submit'])) // name of su...

the code is correct, may be there is some issue with the image location.
so, if only the image name is stored in the database, then the code should be like this:echo "<td><img src=pictures/subfolderI/'".$red['picture_path']."' alt='image' width='150px' height='150px'/></td>";

hello, you can refer this script, hope it works for you : <script type="text/javascript"> function submitform() { document.forms["myform"].submit(); } </script> <!-- give the path where you want to redirect the page and set the method attribute according to that --> <form id="myform" action=""> Text...

Hello, you have used the incorrect method to fetch data from the database; as in this mysql_query the result always obtained in the form of array thats why you are getting Resource Id as an output. To execute this code, apply this way as : <?php $con = mysql_connect("IP.IP.IP.IP","database","passwor...

hello, In the form the method is post and the value of message using with get, so try to use the same whether it is post or get. and in the code try this way, then this notice will not appear <tr> <?php if(isset($_POST['Submit2'])) // if you are using method="post" in form { echo ""; } ?> </tr>

hello, you have missed the header section in the mail function, where we can mention the sender's email id so add these lines to your code : $headers = 'From: <sender's email id>' . "\r\n" . 'X-Mailer: PHP/' . phpversion(); and also mention this variable in the mail function as : mail($MAIL_TO, $sub...

i have tested your script on my system as well and no error in that line, the syntax of if statement is not correct may be because of that, this is not executing, you have missed the else statement after two else if's.